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Question Number 63438 by ajfour last updated on 04/Jul/19

Commented by ajfour last updated on 04/Jul/19

Determine  a:b:c  if the  inscribed △ABC has constant  perimeter but maximum area.

Determinea:b:ciftheinscribedABChasconstantperimeterbutmaximumarea.

Commented by ajfour last updated on 04/Jul/19

MjS Sir, mrW sir, Tanmay Sir,  please try..

MjSSir,mrWsir,TanmaySir,pleasetry..

Answered by ajfour last updated on 04/Jul/19

A(h,h^2 )  B(h+ccos θ, h^2 +csin θ)  C(h+bcos φ, h^2 +bsin φ)  a+b+c=p             ......(i)  h^2 +csin θ=(h+ccos θ)^2        ...(ii)  ⇒  sin θ=ccos^2 θ+2hcos θ  ⇒  ccos θ=tan θ−2h  h^2 +bsin φ=(h+bcos φ)^2       ....(iii)  ⇒  bcos φ=tan φ−2h  (ccos θ−bcos φ)^2 +(csin θ−bsin φ)^2 =a^2                                                      .....(iv)  ⇒ b^2 +c^2 −a^2 =2bccos (φ−θ)  ⇒ −p^2 −2(b+c)p=2bccos (φ−θ)  −p^2 −2p[((tan θ−2h)/(cos θ))+((tan φ−2h)/(cos φ))]         = 2(tan θ−2h)(tan φ−2h)(1+tan θtan φ)  △=((cd)/2)  d=b(sin φ−cos φtan θ)cos θ  ⇒ △=((bc)/2)(bsin φ−bcos φtan θ)cos θ  △=((bc)/2)sin (φ−θ)  2△=(((tan θ−2h)(tan φ−2h)sin (φ−θ))/(cos θcos φ))  2△=(tan θ−2h)(tan φ−2h)(tan φ−tan θ)  −p^2 −2p[((tan θ−2h)/(cos θ))+((tan φ−2h)/(cos φ))]         = 2(tan θ−2h)(tan φ−2h)(1+tan θtan φ)  Now let  tan θ=t,  tan φ=s  ⇒  −p^2 −2p[(t−2h)(√(1+t^2 ))+(s−2h)(√(1+s^2 )) ]     =2(t−2h)(s−2h)(1+st)  but i′m afraid its not going to yield!

A(h,h2)B(h+ccosθ,h2+csinθ)C(h+bcosϕ,h2+bsinϕ)a+b+c=p......(i)h2+csinθ=(h+ccosθ)2...(ii)sinθ=ccos2θ+2hcosθccosθ=tanθ2hh2+bsinϕ=(h+bcosϕ)2....(iii)bcosϕ=tanϕ2h(ccosθbcosϕ)2+(csinθbsinϕ)2=a2.....(iv)b2+c2a2=2bccos(ϕθ)p22(b+c)p=2bccos(ϕθ)p22p[tanθ2hcosθ+tanϕ2hcosϕ]=2(tanθ2h)(tanϕ2h)(1+tanθtanϕ)=cd2d=b(sinϕcosϕtanθ)cosθ=bc2(bsinϕbcosϕtanθ)cosθ=bc2sin(ϕθ)2=(tanθ2h)(tanϕ2h)sin(ϕθ)cosθcosϕ2=(tanθ2h)(tanϕ2h)(tanϕtanθ)p22p[tanθ2hcosθ+tanϕ2hcosϕ]=2(tanθ2h)(tanϕ2h)(1+tanθtanϕ)Nowlettanθ=t,tanϕ=sp22p[(t2h)1+t2+(s2h)1+s2]=2(t2h)(s2h)(1+st)butimafraiditsnotgoingtoyield!

Commented by ajfour last updated on 04/Jul/19

−p^2 −2p[((tan θ−2h)/(cos θ))+((tan φ−2h)/(cos φ))]         = 2(tan θ−2h)(tan φ−2h)(1+tan θtan φ)                                               ........(I)  2△=(tan θ−2h)(tan φ−2h)(tan φ−tan θ)                                               ........(II)  To find θ,φ such that △ is max.

p22p[tanθ2hcosθ+tanϕ2hcosϕ]=2(tanθ2h)(tanϕ2h)(1+tanθtanϕ)........(I)2=(tanθ2h)(tanϕ2h)(tanϕtanθ)........(II)Tofindθ,ϕsuchthatismax.

Commented by mr W last updated on 04/Jul/19

a hard question sir...

ahardquestionsir...

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